Q. When light of wavelength $\lambda$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism for the wavelength $\lambda$ is, then
A) $\sqrt{3}$
B) $\frac{\sqrt{3}}{2}$
C) 2
D) $\sqrt{2}$
Explanation
(a) $\mu = \frac{\sin\left( \frac{A + \delta_{m}}{2} \right)}{\sin\frac{A}{2}} = \frac{\sin\left( \frac{60{^\circ} + 60{^\circ}}{2} \right)}{\sin\left( \frac{60{^\circ}}{2} \right)} = \sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.